For Z=32 , Find the number of completely filled orbitals for which `m_l` = 0

Answer

To solve the problem of finding the number of completely filled orbitals for which \( m_l = 0 \) for an atom with atomic number \( Z = 32 \), we can follow these steps:

### Step 1: Determine the Electron Configuration
The atomic number \( Z = 32 \) corresponds to the element Germanium (Ge). We need to write its electron configuration:
- The electron configuration for \( Z = 32 \) is:
\[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^{10} \, 4s^2 \, 4p^2 \]

### Step 2: Identify the Orbitals and Their \( m_l \) Values
Next, we need to identify the orbitals and their corresponding \( m_l \) values:
- For \( s \) orbitals:
- \( m_l = 0 \) (1 orbital)
- For \( p \) orbitals:
- \( m_l = -1, 0, +1 \) (3 orbitals, but only \( m_l = 0 \) is relevant)
- For \( d \) orbitals:
- \( m_l = -2, -1, 0, +1, +2 \) (5 orbitals, but only \( m_l = 0 \) is relevant)

### Step 3: Check for Completely Filled Orbitals
Now we need to check which of these orbitals are completely filled:
- **1s**: \( 1s^2 \) - completely filled (1 orbital with \( m_l = 0 \))
- **2s**: \( 2s^2 \) - completely filled (1 orbital with \( m_l = 0 \))
- **2p**: \( 2p^6 \) - completely filled (1 orbital with \( m_l = 0 \))
- **3s**: \( 3s^2 \) - completely filled (1 orbital with \( m_l = 0 \))
- **3p**: \( 3p^6 \) - completely filled (1 orbital with \( m_l = 0 \))
- **3d**: \( 3d^{10} \) - completely filled (1 orbital with \( m_l = 0 \))
- **4s**: \( 4s^2 \) - completely filled (1 orbital with \( m_l = 0 \))
- **4p**: \( 4p^2 \) - not completely filled (does not contribute)

### Step 4: Count the Completely Filled Orbitals with \( m_l = 0 \)
Now, we count the completely filled orbitals where \( m_l = 0 \):
- 1s: 1
- 2s: 1
- 2p: 1
- 3s: 1
- 3p: 1
- 3d: 1
- 4s: 1

Adding these up gives:
\[ 1 + 1 + 1 + 1 + 1 + 1 + 1 = 7 \]

### Final Answer
The number of completely filled orbitals for which \( m_l = 0 \) is **7**.

---